similar to: Must be obvious but not to me : problem with regular expression

Hi, I recently discovered the R program and I thought it could be useful to me. I have to analyse data saved as .Px file (x between 0 and 8 - .P0 files have 18 lines at the beginning that I have to skip). New files are generated everyday. This is my strategy : In order to analyse the data, I first want to copy the new data in a database in MySQL (which already contains the previous data). So

Dear All,   I was wondering if you could help me with the following: I have the code:   tin <-0.5 tau <-24 output0 <-10 TIMELOW <-tin TIMEHIGH <-1*tau TIME1 <-c(seq(TIMELOW,TIMEHIGH, by = sign(TIMEHIGH-TIMELOW)*(tau-tin)/3))   then I would like to calculate:   cp1 <-output0*exp(-0.3*TIME1[1]) cp2 <-output0*exp(-0.3*TIME1[2]) cp3 <-output0*exp(-0.3*TIME1[3]) cp4

Hi again, This is the follow of my post "Problem with save or/and if (I think but maybe not ...)". In this post, I wrote that I solved my main problem. And it is true. I also wrote that there was still another problem, which I managed to solve. But I think there must be another way to solve it taking advantages of the R language (which I don't master at all), that is with less

Le 29 mai 2010 à 01:08, Bill Wendling a écrit : > Hi Stéphane, > > The SSE support is the LLVM backend is fine. What is the code that's generated? Do you have some short examples of where LLVM doesn't do as well as the equivalent scalar code? > > -bw > > On May 28, 2010, at 12:13 PM, Stéphane Letz wrote: We are actually testing LLVM for the Faust language

On Sat, May 29, 2010 at 12:42 AM, Stéphane Letz <letz at grame.fr> wrote: > > Le 29 mai 2010 à 01:08, Bill Wendling a écrit : > >> Hi Stéphane, >> >> The SSE support is the LLVM backend is fine. What is the code that's generated? Do you have some short examples of where LLVM doesn't do as well as the equivalent scalar code? >> >> -bw >>

On Sat, May 29, 2010 at 1:18 AM, Eli Friedman <eli.friedman at gmail.com> wrote: > On Sat, May 29, 2010 at 12:42 AM, Stéphane Letz <letz at grame.fr> wrote: >> >> Le 29 mai 2010 à 01:08, Bill Wendling a écrit : >> >>> Hi Stéphane, >>> >>> The SSE support is the LLVM backend is fine. What is the code that's generated? Do you have some

Hello, I'm still trying to make the life of my colleagues easier. Nice, isn't it ? At the moment, I'm looking for a way to generate multiple "single report". In fact I have a .rnw file which send a query to a MySQL database (rs<-dbSendQuery(con, statement="select * from treatdata where name='Device1'") But of course my colleagues have many devices and

Hi, I have a data frame of 2 columns with the following types : data$day char data$value num And I plot my data with : plot(strptime(donnees$day,format="%Y-%m-%d %H:%M:%S"),donnees$value, type="l") And I'd just like to get the numerical value of the integration of this graph. I looked at ?integrate but, as far as I understood (that is, not very much, due to my poor

Hi Stéphane, The SSE support is the LLVM backend is fine. What is the code that's generated? Do you have some short examples of where LLVM doesn't do as well as the equivalent scalar code? -bw On May 28, 2010, at 12:13 PM, Stéphane Letz wrote: > Hi, > > We are experimenting directly generating vectorized LLVM IR (using <8 x float> kind of types), then compiling the code

Hi, We are experimenting directly generating vectorized LLVM IR (using <8 x float> kind of types), then compiling the code to SSE on a 64 bits machine. Right now the equivalent code in scalar mode sill outperform the SSE one. What is the quality of the SSE support in X86 LLVL backend? Are they any specific things to be aware of to improve the speed? Thanks Stéphane Letz

Hello, When I plot y=f(x) from the file xy.txt ( http://www.nabble.com/file/p18773387/xy.txt xy.txt ), I can clearly see a trend. Is there a function or a package able to take the median value of y for an interval of x (x +/- a defined value) to plot nice graph (at least a better one) ? Thanks in advance, Have a nice week-end, Ptit Bleu. -- View this message in context:

Hello, I'm just looking for an easy way to find the positions of a complete sequence in a bigger vector. For example : c("a","z","e") in c("a","z","e","r","t","a","z","a","z","e","c") and the result should be 1 8 that is the positions of the beginning of

Thanks to your help, I managed to link R and a Mysql Database, send queries, plot the results and put everything in a pdf document (with Sweave). My co-workers find the job not bad but they would like to have a friendly interface to send queries and see the graphs (the pdf document is not necessary). Something like this coul be a good starting point : --------------------------------------- |

Hi, I read the posts for 2 hours and ?list and tried many comninations but I haven't found the answer to this basic question. So I decided to post my question even if it is a silly one ... What is the instruction to retrieve, for example, the "D" of the first list ? Thanks in advance, Ptit Bleu. > x<-list(LETTERS[1:5], LETTERS[10:20]) > x [[1]] [1] "A"

Hello, I have a text file with one column containing long number but stored as string. I download the file with read.table (and colClass) and the first row of this column is : "095842087016731010" As I need to make some calculations with these numbers, I tried to convert them using as.numeric. But then I get as.numeric("095842087016731010") 95842087016731008 I understand

Hi, I keep on trying to write some small scripts in order to learn R but even with basic scripts I have problems ... I start with the name of a file which is in fact the time the file has been generated (I cannot change the format). Then I convert namefile with strptime. The problem occurs when I add another time from another file with append. It displays some informations I don't want. I

Hello, I have a list of data.frame rowsplit : List of 15 $ (0,0.025] :'data.frame': 169 obs. of 7 variables: $ (0.025,0.05]:'data.frame': 174 obs. of 7 variables: $ (0.05,0.075]:'data.frame': 92 obs. of 7 variables: $ (0.075,0.1] :'data.frame': 76 obs. of 7 variables: $ (0.1,0.125] :'data.frame': 37 obs. of 7 variables: $

Hello, I would like to fit data with the following formula : y=V*(1+alpha*(x-25)) where y and x are my data, V is a constant and alpha is the slope I'm looking for. How to translate this into R-language ? At the moment, I only know : lm(y ~ x) Sorry for such a basic question. I thought I could find the solution in a post but I have to confess that, up to know, I'm not able to understand

Hello, I would like to replace a for loop with lapply in order to speed up the treatment of my data (I've read lapply can help to do it). At the end of the message, you will find a simple example (function is more complex and data frames is more than 1 million of rows) of what I would like to do, that is applying the same calculation to all the rows of the data frame (knowing that I cannot

Hello, I'm trying to fit some points with a 8-degrees polynom (result of lm is stored in pfit). In most of the case, it is ok but for some others, some coefficients are "NA". I don't really understand the meaning of these "NA". And the problem is that I can't perform a derivation (pderiv<-as.function((deriv(polynomial(pfit$coefficients))))) on pfit due to the